Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 12}{x + 2} = \dfrac{5x + 26}{x + 2}$
Multiply both sides by $x + 2$ $ \dfrac{x^2 + 12}{x + 2} (x + 2) = \dfrac{5x + 26}{x + 2} (x + 2)$ $ x^2 + 12 = 5x + 26$ Subtract $5x + 26$ from both sides: $ x^2 + 12 - (5x + 26) = 5x + 26 - (5x + 26)$ $ x^2 + 12 - 5x - 26 = 0$ $ x^2 - 14 - 5x = 0$ Factor the expression: $ (x + 2)(x - 7) = 0$ Therefore $x = -2$ or $x = 7$ At $x = -2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -2$, it is an extraneous solution.